8 Applications of Cauchy’s theorem

Theorem 8.1 (Cauchy’s integral formula) Let $f\colon U\to \C$ be a holomorphic function and assume that $\ol D_r(z_0)\subset U$ with boundary curve \[[0,2\pi]\xrightarrow{\ga_{\6 D_r(z_0)}}\ol D_r(z_0), \ga_{\6 D_r(z_0)}(t) = z_0+re^{it}.\] Then

\[f(z) = \frac{1}{2\pi i}\int_{\6D_r(z_0)}\frac{f(\ze)}{\ze-z}d\ze,\qquad\forall z\in D_r(z_0). \tag{8.1}\]

Proof.

Fix $z\in D_r(z_0).$ Since $f$ is complex differentiable at $z,$ we have \[f(\ze)=f(z)+(\ze-z)f'(z)+(\ze-z)\rho(\ze-z),\lim_{\ze\to z}\rho(\ze-z)=0. \tag{8.2}\]

There is a homotopy in $U\setminus\{z\}$ between the loops $\6D_r(z_0)$ and $\6D_s(z),$ for all $s>0$ with $D_s(z)\subset D_r(z_0),$ see Figure 8.1.

Figure 8.1: Homotopy between $\6D_r(z_0)$ and $\6D_s(z)$

Moreover, $\ze\mapsto\frac{f(\ze)}{\ze-z}$ is a holomorphic function on $U\setminus\{z\}.$

By Theorem 7.1 we have \[\int_{\6D_r(z_0)}\frac{f(\ze)}{\ze-z}d\ze = \int_{\6D_s(z)}\frac{f(\ze)}{\ze-z}d\ze,\] applying Equation 8.2, \[\int_{\6D_s(z)}\frac{f(\ze)}{\ze-z}d\ze = \int_{\6D_s(z)}\frac{f(z)}{\ze-z}d\ze +\int_{\6D_s(z)}f'(z)d\ze+\int_{\6D_s(z)}\rho(\ze-z)d\ze\] applying Equation 6.6 first with $n=-1$ and then $n =0$, we have \[\begin{align} \int_{\6D_s(z)}\frac{f(z)}{\ze-z}d\ze +\int_{\6D_s(z)}f'(z)d\ze+\int_{\6D_s(z)}\rho(\ze-z)d\ze &= 2\pi if(z) \\ &\phantom{+} +0 \\ &\phantom{+} +\int_{\6D_s(z)}\rho(\ze-z)d\ze. \end{align}\]

Consequently, \[\int_{\6D_r(z_0)}\frac{f(\ze)}{\ze-z}d\ze= 2\pi if(z) +0 +\int_{\6D_s(z)}\rho(\ze-z)d\ze.\]

Since $\lim_{\ze\to z}\rho(\ze-z)=0$ there exists a bound $|\rho(\ze-z)|\leqslant M.$ Hence the final term can be estimated using Equation 6.7 as \[\left|\int_{\6D_s(z)}\rho(\ze-z)d\ze\right|\leqslant 2\pi sM.\]

This tends to zero as $s\to 0,$ and the left hand side is independent of $s.$

Theorem 8.2 (Higher Cauchy integral formula) Let $f\colon U\to \C$ be a holomorphic function and assume that $\ol D_r(z_0)\subset U.$ Then $f$ is equal on $D_r(z_0)$ to its Taylor power series

\[f(z)=\sum_{n=0}^\iy\frac{f^{(n)}(z_0)}{n!}(z-z_0)^n, \tag{8.3}\]

which has radius of convergence $\rho\geqslant r.$ In particular, $f$ is infinitely complex differentiable on the open set $U$ and has a primitive on $D_r(z_0).$ Moreover,

\[f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_{\6 D_r(z_0)}\frac{f(\ze)}{(\ze-z_0)^{n+1}}d\ze. \tag{8.4}\]

Proof.

Let $z\in D_r(z_0).$ The integrand in Equation 8.1 can be rewritten as

\[\frac{f(\ze)}{\ze-z}=\frac{\frac{f(\ze)}{\ze-z_0}}{1-\frac{z-z_0}{\ze-z_0}} =\frac{f(\ze)}{\ze-z_0}\sum_{n=0}^\iy\left(\frac{z-z_0}{\ze-z_0}\right)^n, \tag{8.5}\]

where the geometric series converges uniformly for all $\ze$ with $|z-z_0|<|\ze-z_0|.$

The region $|z-z_0|<|\ze-z_0|$ and the contour of integration

In particular, this holds for all $\ze\in\6D_r(z_0)$ and using Corollary 6.1 we may exchange the limit and the integral. Hence for all $|z-z_0|<r$ we have a convergent series

\[f(z) = \sum_{n=0}^\iy\left(\frac{1}{2\pi i}\int_{\6 D_r(z_0)}\frac{f(\ze)}{(\ze-z_0)^{n+1}}d\ze\right)(z-z_0)^n. \tag{8.6}\]

In particular, the series Equation 8.6 has radius of convergence $\rho\geqslant r.$

Finally, Equation 8.4 (and therefore Equation 8.3) follow from the fact that the power series Equation 8.6 is differentiated termwise, see Theorem 5.2.

Theorem 8.3 (Liouville) Every bounded entire function is constant.

Proof.

Suppose $|f(z)|<C$ for all $z\in\C.$ Then by Equation 8.4 with $z_0=0$ we have \[|f^{(n)}(0)|=\left|\frac{n!}{2\pi i}\int_{\6D_r(0)}\frac{f(\ze)}{\ze^{n+1}}d\ze\right|,\] by Equation 6.7, \[\left|\frac{n!}{2\pi i}\int_{\6D_r(0)}\frac{f(\ze)}{\ze^{n+1}}d\ze\right|\leqslant \frac{n!}{2\pi}L(\6D_r(0))\frac{C}{r^{n+1}}=\frac{n!C}{r^n}\] and so \[|f^{(n)}(0)| \leqslant \frac{n!C}{r^n}\] for all $r>0.$ Letting $r\to\iy,$ this implies that $f^{(n)}(0)=0$ for all $n>0.$ Now Equation 8.3 proves that $f(z)$ is constant.

Carl Friedrich Gauß, 1777-1855, [Österreichische Nationalbibliothek](http://data.onb.ac.at/rec/baa8260342). Public Domain — Carl Friedrich Gauß, 1777-1855, Österreichische Nationalbibliothek. Public Domain

Theorem 8.4 (Fundamental theorem of algebra) Every non-constant polynomial $P(z)=a_nz^n+\ldots+a_1z+a_0$ with $a_i\in\C$ has a complex root.

Proof.

Assume by contradiction that $P(z)\neq0$ for all $z\in\C.$ Then $1/P(z)$ is holomorphic and bounded. Indeed, assuming $a_n\neq0,$ we find \[\begin{align*} |P(z)|\geqslant |a_n||z|^n-|a_{n-1}||z|^{n-1}-\ldots-|a_0|\to+\iy \text{as $z\to\iy.$} \end{align*}\] Hence $1/|P(z)|\to0$ as $z\to\iy.$ In particular, $|1/P(z)|$ is bounded.

Definition 8.1

A subset $D\subset\C$ is path-connected if for all $z_0, z_1\in D$ there exists a (piecewise C¹) curve $\ga$ in $D$ with $\ga(0)=z_0,$ $\ga(1)=z_1.$
A path-connected subset $D$ is simply connected if every loop in $D$ is (freely) homotopic in $D$ to a constant loop.

Example 8.1

Every star-shaped set $D$ is simply connected. Firstly, $D$ is path-connected since any $z\in D$ can be connected by a straight line $tz+(1-t)z_0$ to the focal point $z_0.$ Secondly, if $\ga\colon[a,b]\to D$ is a loop, then $\Ga_s(t)=sz_0+(1-s)\ga(t)$ is a homotopy in $D$ from $\ga$ to the constant loop.

Theorem 8.5 Let $f\colon U\to\C$ be a holomorphic function. Suppose that $U$ is simply connected and let $z_0\in U.$ Define $F(z)$ for each $z\in U$ by choosing a piecewise C¹ curve $\ga$ with $\ga(0)=z_0,$ $\ga(1)=z$ and defining

\[F(z)=\int_\ga f(\ze)d\ze. \tag{8.7}\]

Then $F$ is well-defined and is the unique holomorphic function on $U$ such that $F'=f$ and $F(z_0)=0.$

Proof.

For any two curves $\ga_0, \ga_1$ as in the statement of the theorem, the concatenation $\ga_0\ast(-\ga_1)$ is a loop which is null-homotopic by assumption. By Theorem 7.2 we have \[\int_{\ga_0} f(\ze)d\ze - \int_{\ga_1}f(\ze)d\ze = 0,\]

hence Equation 8.7 is well-defined. It remains to check that $F$ is complex differentiable at every point $z_*\in U$ for which it suffices to restrict attention to the disk $\ol D_r(z_*)\subset U.$

Fix a path $\ga_*$ in $U$ from $z_0$ to $z_*.$ For each $z\in D_r(z_*),$ let $\ga$ be the concatenation of $\ga_*$ with a path $\eta$ from $z_*$ to $z$ in the disk $D_r(z_*).$ Then \[F(z)=\int_\ga f(\ze)d\ze=\int_{\ga_*}f(\ze)d\ze+\int_\eta f(z)d\ze.\]

The first integral is a constant independent of $z.$ By Theorem 8.2, $f$ has a holomorphic primitive $g$ on $D_r(z_*)$ and $\eta$ is a path in that disk, so we can apply the complex FTC Equation 6.8 to the second integral and write \[F(z)=\int_{\ga_*}f(\ze)d\ze+g(z)-g(z_*).\]

This shows that $F$ is a holomorphic function of $z$ with $F'=g'=f.$

To prove uniqueness, suppose that $\tilde F$ is another holomorphic function with $\tilde F'=f$ and $\tilde F(z_0)=w_0.$ Then $F-\tilde F=c$ is constant on $U$ and by evaluating at $z_0$ we find $c=F(z_0)-\tilde F(z_0)=w_0-w_0=0.$

Example 8.2

Let $U\subset\C$ be a simply connected subset with $0\notin U$ and $f(z)=1/z.$ Fix $w_0\in\C$ such that $z_0=e^{w_0}\in U.$ Define the holomorphic function $F(z)$ on $U$ by Equation 8.7. Then $\ell(z)=F(z)+w_0$ is the unique branch of the logarithm on $U$ such that $\ell(z_0)=w_0.$

For $z_0=1,$ $w_0=0,$ and $U=\C^-,$ we recover the principal branch of the logarithm, since $\log'(z)=1/z$ by Equation 4.9 and $\log(1)=0.$

Questions for further discussion

How should $\int_{-i}^{1+i} z^2dz$ be interpreted? What is the result?
Why is $\C^\t$ path-connected but not simply connected?

Hint: consider $\frac{1}{2\pi i}\int_\ga \frac{1}{z}dz$ for a loop $\ga$ in $\C^\t.$ +
It is a mysterious calculus fact that $\arctan(x)=\sum_{n=0}^\iy (-1)^n\frac{x^{2n+1}}{2n+1}$ diverges for $|x|> 1$ although $\arctan\colon\R\to\R$ is smooth. In terms of the principal logarithm, $\arctan(z)=\frac{i}{2}\log\left(\frac{i+z}{i-z}\right).$ Apply this to give a geometric explanation of the divergence using Theorem 8.2.

8.1 Exercises

Exercise 8.1

Compute the following integrals:

$\int_{\partial D_1(0)}\frac{\sin(z)}{z^2}dz,$
$\int_{\partial D_{1/2}(0)}\frac{\cos(z)}{z-1}dz,$
$\int_{\partial D_2(1)}\frac{\sin(\cos(z))}{z-1}dz.$

Exercise 8.2

Let $f\colon\C\to\C$ be entire. Suppose there is a constant $C>0$ such that $|f(z)|\leqslant C|z|^d$ for all $z\in\C.$ Prove that $f(z)$ is a polynomial of degree $\leqslant d.$

Hint: Generalize the proof of Liouville’s theorem.

Exercise 8.3

Let $f\colon U\to\C$ be holomorphic and $\ol D_r(z_0)\subset U.$ Show that \[f(z_0)=\frac{1}{2\pi}\int_0^{2\pi} f(z_0+re^{i\varphi})d\varphi\]

and interpret this equation geometrically.

Exercise 8.4

Let $f\colon U\to\C$ be holomorphic and $\ol D_r(z_0)\subset U.$

Prove that $g(x)=f(z_0+re^{ix})$ is a $2\pi$-periodic function $\R\to\C.$
Show that the Cauchy integral formula implies an absolutely and uniformly convergent Fourier expansion \[g(x)=\sum_{n=0}^\iy \ga_ne^{inx},\qquad\forall x\in\R,\] with only non-negative Fourier modes. Moreover, show that $\ga_n=\frac{1}{2\pi}\int_0^{2\pi}g(x)e^{-inx}dx.$$

Note. More generally, a bi-infinite Fourier series $\sum_{n=-\iy}^{\iy}\ga_n e^{inx}$ can be obtained as a superposition $g_1(x)+g_2(-x).$

Exercise 8.5

Let $f\colon U\to\C$ be holomorphic and $\ol D_r(z_0)\subset U.$ Using polar coordinates show that for the (surface) integral over the unit disk $\ol D_r(x_0)=\{(x,y)\in\R^2\mid (x-x_0)^2+(y-y_0)^2\leqslant r\}$ we have \[f(z_0)=\frac{1}{\pi r^2}\int_{\ol D_r(z_0)}f(x+iy)dxdy\]

and interpret this equation geometrically.